Open mapping theorem (functional analysis)

Posted on February 24, 2020 by lordneo

Suppose

{displaystyle A:Xto Y}

${displaystyle A:Xto Y}$ is a surjective continuous linear operator. In order to prove that

{displaystyle A}

$A$ is an open map, it is sufficient to show that

{displaystyle A}

$A$ maps the open unit ball in

{displaystyle X}

$X$ to a neighborhood of the origin of

{displaystyle Y.}

$Y.$

Let

{displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}

${displaystyle U=B_{1}^{X}(0),V=B_{1}^{Y}(0).}$
Then

{displaystyle X=bigcup _{kin mathbb {N} }kU.}

Since

{displaystyle A}

$A$ is surjective:

{displaystyle Y=A(X)=Aleft(bigcup _{kin mathbb {N} }kUright)=bigcup _{kin mathbb {N} }A(kU).}

But

{displaystyle Y}

$Y$ is Banach so by Baire’s category theorem

{displaystyle exists kin mathbb {N} :qquad left({overline {A(kU)}}right)^{circ }neq varnothing .}

That is, we have

{displaystyle cin Y}

${displaystyle cin Y}$ and

{displaystyle r>0}

$r > 0″/></span> such that </p> <div class=$

${displaystyle B_{r}(c)subseteq left({overline {A(kU)}}right)^{circ }subseteq {overline {A(kU)}}.}$

Let

{displaystyle vin V,}

${displaystyle vin V,}$ then

{displaystyle c,c+rvin B_{r}(c)subseteq {overline {A(kU)}}.}

By continuity of addition and linearity, the difference

{displaystyle rv}

${displaystyle rv}$ satisfies

{displaystyle rvin {overline {A(kU)}}+{overline {A(kU)}}subseteq {overline {A(kU)+A(kU)}}subseteq {overline {A(2kU)}},}

and by linearity again,

{displaystyle Vsubseteq {overline {A(LU)}}}

where we have set

{displaystyle L=2k/r.}

${displaystyle L=2k/r.}$
It follows that for all

{displaystyle yin Y}

$yin Y$ and all

{displaystyle epsilon >0,}

$epsilon > 0,”/></span> there exists some <span class=$

{displaystyle xin X}

$xin X$ such that

{displaystyle qquad |x|_{X}leq L|y|_{Y}quad {text{and}}quad |y-Ax|_{Y}

Our next goal is to show that

{displaystyle Vsubseteq A(2LU).}

${displaystyle Vsubseteq A(2LU).}$

Let

{displaystyle yin V.}

$yin V.$
By (1), there is some

{displaystyle x_{1}}

$x_{1}$ with

{displaystyle left|x_{1}right|

${displaystyle left|x_{1}right|<L}$ and

{displaystyle left|y-Ax_{1}right|<1/2.}

${displaystyle left|y-Ax_{1}right|<1/2.}$
Define a sequence

{displaystyle left(x_{n}right)}

${displaystyle left(x_{n}right)}$ inductively as follows.
Assume:

{displaystyle |x_{n}|<{frac {L}{2^{n-1}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)right|<{frac {1}{2^{n}}}.qquad (2)}

Then by (1) we can pick

{displaystyle x_{n+1}}

$x_{n+1}$ so that:

{displaystyle |x_{n+1}|<{frac {L}{2^{n}}}quad {text{and}}quad left|y-Aleft(x_{1}+x_{2}+cdots +x_{n}right)-Aleft(x_{n+1}right)right|<{frac {1}{2^{n+1}}},}

so (2) is satisfied for

{displaystyle x_{n+1}.}

${displaystyle x_{n+1}.}$ Let

{displaystyle s_{n}=x_{1}+x_{2}+cdots +x_{n}.}

From the first inequality in (2),

{displaystyle left(s_{n}right)}

${displaystyle left(s_{n}right)}$ is a Cauchy sequence, and since

{displaystyle X}

$X$ is complete,

{displaystyle s_{n}}

$s_{n}$ converges to some

{displaystyle xin X.}

${displaystyle xin X.}$
By (2), the sequence

{displaystyle As_{n}}

${displaystyle As_{n}}$ tends to

{displaystyle y}

$y$ and so

{displaystyle Ax=y}

$Ax=y$ by continuity of

{displaystyle A.}

$A.$
Also,

{displaystyle |x|=lim _{nto infty }|s_{n}|leq sum _{n=1}^{infty }|x_{n}|<2L.}

This shows that

{displaystyle y}

$y$ belongs to

{displaystyle A(2LU),}

${displaystyle A(2LU),}$ so

{displaystyle Vsubseteq A(2LU)}

${displaystyle Vsubseteq A(2LU)}$ as claimed.
Thus the image

{displaystyle A(U)}

$A(U)$ of the unit ball in

{displaystyle X}

$X$ contains the open ball

{displaystyle V/2L}

${displaystyle V/2L}$ of

{displaystyle Y.}

$Y.$
Hence,

{displaystyle A(U)}

$A(U)$ is a neighborhood of the origin in

{displaystyle Y,}

$Y,$ and this concludes the proof.